Gravitation and orbital motion
Gravitation and orbital motion
This domain tests Newton's universal gravitation, gravitational field and potential, satellite motion, Kepler's laws and escape velocity. Questions often ask you to compare a low-orbit satellite with a geostationary one, or to relate g at altitude to g at Earth's surface.
Core rules
- Newton's law: F=GMm/r², attractive and along the line joining centres. Gravitational field is g=GM/r², directed toward the mass.
- Potential energy: U=−GMm/r; the zero is usually at infinity, so bound orbits have negative total energy. Potential is V=−GM/r.
- Circular orbits: gravity supplies the centripetal force, GMm/r²=mv²/r, so v=√(GM/r). Orbital speed does not depend on the satellite's mass.
- Periods and Kepler: T=2πr/v gives T²=4π²r³/(GM). Kepler's third law applies to any central mass, with r the semi-major axis for elliptical orbits.
- Geostationary orbit: above the equator with a 24 h period, so the satellite stays over one point; its orbital radius is about 4.2×10⁷ m from Earth's centre.
- Escape velocity: v_esc=√(2GM/R), independent of launch angle and mass.
Worked example
Find the speed and period of a satellite in a circular orbit at r=2R_E, with R_E=6.37×10⁶ m and g=9.8 m/s². Use GM=gR_E². Then v=√(gR_E²/(2R_E))=√(gR_E/2)=√(9.8×6.37×10⁶/2)=5.6×10³ m/s. The period is T=2πr/v=2π(1.274×10⁷)/(5.59×10³)=1.43×10⁴ s, about 4.0 h. Note that r is measured from Earth's centre, not from the surface.
Traps
Substituting altitude for r; thinking a more massive satellite needs a different orbital speed; using positive gravitational potential energy; applying Kepler's law with the satellite mass instead of the central mass; confusing escape velocity with orbital velocity.